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3 years ago

A bar magnet is held in place while another bar magnet is placed near it. The second bar magnet spins around and

Physics

1 answer:

Vinil7 [7]3 years ago

4 0

Answer:

Explanation:

the increase energy stored in thw system is proportional to the decrease in kinetic energy

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A leopard of mass 65kg climbs 7m up a tall tree. Calculate how much gravitational potential energy it gains. Assume g=10N/kg.

vesna_86 [32]

Answer: 4550 Joules

The formula for potential energy in gravitational field is as follows:

$E_{pot} = mgh=65kg \cdot 10 \frac{N}{kg} \cdot 7m = 4550 J$

8 0

3 years ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0Â° east of

koban [17]

<u>Answer:</u>

Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

Speed of truck = 25 m/s north = 25 j m/s

Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

= (1.43 i + 1.00 j) m/s

Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

Magnitude of velocity = 26.04 m/s

Angle from positive horizontal axis = 86.85⁰

So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0

3 years ago

50 points!!! Kinetics

iris [78.8K]

Answer:

98 m √

Explanation:

How about s = Vo * t + ½at² ?

s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6

and

h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490

Subtract 2nd from first:

0 = -8Vo + 470.4

Vo = 58.8 m/s

h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m

3 0

2 years ago

Read 2 more answers

An unknown gas fills a cubical container of 0.87 m. If the mass of the gas is 0.316 g, what is the density of the gas ?

svetlana [45]

Answer:

The length of the Mercury column of thermometer at ice point is 20mm and 220mm at steam point . when the same Thermometer is placed in contact with another body ,it reads 5°c.what will the length of the Mercury column at the temperature?

3 0

1 year ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

7 0

3 years ago